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• TI Thinks Resolved

# LM239: lm239

Part Number: LM239

Good morning, I am using LM239 and I want to supply it with 11V whereas my output pull-up voltage is 35V tied to 15k pull-up resistors. Which is the maximum output voltage of LM239 when supplied with 11V?

Thank you!

• Hi Roger,

I would not go beyond what is recommended in section 7.3 of datasheet. 35V is all too close to the absolute maximum rating of 36V, potentially leaving you without enough reserve.

Kai

• In reply to kai klaas69:

Good morning thank you for the reply. So the pull up voltage is independent from the supply? So If I supply with 11V it is possible to have higher voltage to Vout (es.24V or 35V )? Thank you.
• In reply to ROGER BORZUMATI:

Hi Roger,

yes, that's what section 9.1 of datasheet is saying.

Kai
• In reply to kai klaas69:

Hi, ok..I will supply LM239 to Vsup= 11V and my 15k pull-up output resistor will be supplied by a higher voltage Vlogic= 24V or 35V(considering that with this voltage and will not have enough margin..). Thank you.
• In reply to kai klaas69:

Thank you for the reply.. I have another question.. What about the input common mode range of the comparator.. It is valid what the datasheet says about out of range at page 13?:

The following list describes the outcomes of some input voltage situations.

• When both IN– and IN+ are both within the common-mode range:

– If IN– is higher than IN+ and the offset voltage, the output is low and the output transistor is sinking

current

– If IN– is lower than IN+ and the offset voltage, the output is high impedance and the output transistor is

not conducting

• When IN– is higher than common mode and IN+ is within common mode, the output is low and the output

transistor is sinking current

• When IN+ is higher than common mode and IN– is within common mode, the output is high impedance and

the output transistor is not conducting

• When IN– and IN+ are both higher than common mode, the output is low and the output transistor is sinking

current

Thank you..

• In reply to ROGER BORZUMATI:

Hi Roger,

yes, this is correct.

Kai
• In reply to kai klaas69:

If you have any additional questions, feel free to re-open the thread by hitting the Reply. Otherwise, selecting Resolved would be appreciated.
Thanks
Chuck

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